What is the integral of (\sqrt[3]{x-2})/(x-3) ?

The integral of (\sqrt[3]{x-2})/(x-3) is

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integral of (\sqrt[3]{x-2})/(x-3)

Solution steps

Apply u-substitution

: \int \frac{3 v ^{3}}{v ^{3} - 1} d v

= \int \frac{3 v ^{3}}{v ^{3} - 1} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{v ^{3}}{v ^{3} - 1} d v

Take the partial fraction of

\frac{v ^{3}}{v ^{3} - 1} : 1 + \frac{1}{3 ( v - 1 )} + \frac{- v - 2}{3 ( v ^{2} + v + 1 )}

= 3 \cdot \int 1 + \frac{1}{3 ( v - 1 )} + \frac{- v - 2}{3 ( v ^{2} + v + 1 )} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int 1 d v + \int \frac{1}{3 ( v - 1 )} d v + \int \frac{- v - 2}{3 ( v ^{2} + v + 1 )} d v)

\int 1 d v = v

\int \frac{1}{3 ( v - 1 )} d v = \frac{1}{3} ln ∣ v - 1 ∣

\int \frac{- v - 2}{3 ( v ^{2} + v + 1 )} d v = \frac{1}{6 3} (- 3 ln 4 v^{2} + 4 v + 4 - 6 arctan (\frac{2 v + 1}{3}))

= 3 (v + \frac{1}{3} ln ∣ v - 1 ∣ + \frac{1}{6 3} (- 3 ln 4 v^{2} + 4 v + 4 - 6 arctan (\frac{2 v + 1}{3})))

Substitute back

Simplify

Add a constant to the solution

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What is the integral of (\sqrt[3]{x-2})/(x-3) ?
The integral of (\sqrt[3]{x-2})/(x-3) is